Problem 36 - Algebra and Trigonometry

Question

Use Cramer’s Rule to solve the system. $$ \left\\{\begin{array}{l}{\frac{1}{2} X+\frac{1}{3} y=1} \\ {\frac{1}{4} x-\frac{1}{6} y=-\frac{3}{2}}\end{array}\right. $$

Step-by-Step Solution

Answer

The solution is \( x = -1 \) and \( y = 6 \).

Step 1: Identify the Coefficients

The given system of equations is: \( \frac{1}{2}x + \frac{1}{3}y = 1 \) and \( \frac{1}{4}x - \frac{1}{6}y = -\frac{3}{2} \). Identify the coefficients of \( x \) and \( y \) which are: \( a = \frac{1}{2}, b = \frac{1}{3}, c = \frac{1}{4}, d = -\frac{1}{6} \).

Step 2: Write the Determinant of the Coefficient Matrix

Calculate the determinant of the coefficient matrix \( D \). The matrix is: \[\begin{pmatrix}\frac{1}{2} & \frac{1}{3} \\frac{1}{4} & -\frac{1}{6} \\end{pmatrix}\]The determinant \( D \) is calculated as:\[ D = \left( \frac{1}{2} \times -\frac{1}{6} \right) - \left( \frac{1}{3} \times \frac{1}{4} \right) \]

Step 3: Calculate Determinant D

Now calculate the determinant \( D \): \[ D = \left( \frac{1}{2} \times -\frac{1}{6} \right) - \left( \frac{1}{3} \times \frac{1}{4} \right) = -\frac{1}{12} - \frac{1}{12} \ = -\frac{2}{12} = -\frac{1}{6} \]

Step 4: Determine Determinant D_x

Replace the first column of the coefficient matrix with the constants from the right side of the equations to form matrix \( D_x \):\[\begin{pmatrix}1 & \frac{1}{3} \-\frac{3}{2} & -\frac{1}{6} \\end{pmatrix}\]Calculate determinant \( D_x \):\[ D_x = \left(1 \times -\frac{1}{6} \right) - \left( \frac{1}{3} \times -\frac{3}{2} \right) \]

Step 5: Calculate Determinant D_x

Calculate determinant \( D_x \):\[ D_x = -\frac{1}{6} + \frac{1}{2} = \frac{2}{12} = \frac{1}{6} \]

Step 6: Determine Determinant D_y

Replace the second column of the coefficient matrix with the constants from the right side of the equations to form the matrix \( D_y \):\[\begin{pmatrix}\frac{1}{2} & 1 \\frac{1}{4} & -\frac{3}{2} \\end{pmatrix}\]Calculate determinant \( D_y \):\[ D_y = \left( \frac{1}{2} \times -\frac{3}{2} \right) - \left( 1 \times \frac{1}{4} \right) \]

Step 7: Calculate Determinant D_y

Calculate the determinant \( D_y \):\[ D_y = -\frac{3}{4} - \frac{1}{4} = -1 \]

Step 8: Solve for x and y using Cramer's Rule

Using Cramer's Rule, the solutions are given by \( x = \frac{D_x}{D} \) and \( y = \frac{D_y}{D} \).Calculate \( x = \frac{\frac{1}{6}}{-\frac{1}{6}} = -1 \).Calculate \( y = \frac{-1}{-\frac{1}{6}} = 6 \).

Key Concepts

Systems of Equations

A system of equations consists of two or more equations with the same variables. The main goal when dealing with systems of equations is to find values for the variables that satisfy all the equations simultaneously. In our original exercise, we were given two linear equations involving two variables, \(x\) and \(y\). Each equation represents a line on a graph, and the solution is the point where these two lines intersect.
Common methods to solve systems of equations include substitution, elimination, and using Cramer's Rule, which is useful for smaller systems, particularly those with two or three variables. Cramer's Rule is a deterministic approach that uses determinants from linear algebra to find the point of intersection without graphing. It's important to ensure the system has a unique solution, which is typically when the determinant of the coefficient matrix is non-zero.

Determinants

In linear algebra, determinants are a special number associated with matrices. For a system of linear equations, the determinant helps determine whether the equations have a unique solution.
In our solution, we calculate the determinant of the coefficient matrix, essential for applying Cramer's Rule. The determinant \(D\) of a \(2 \times 2\) matrix \(\begin{pmatrix} a & b \ c & d \end{pmatrix}\) is given by: \[ D = ad - bc \] This determinant is crucial because if it equals zero, the system has either no solutions or an infinite number of solutions. During Step 3 of the exercise, we calculated the determinants \(D\), \(D_x\), and \(D_y\), which allowed us to find the values of \(x\) and \(y\) using Cramer's Rule. The ability to calculate these determinants is a fundamental skill in linear algebra and is often used in solving network problems, transforming spaces, and changing coordinate systems.

Linear Algebra

Linear algebra is a branch of mathematics that deals with vectors, vector spaces, and linear equations. It plays a crucial role in various fields like computer graphics, physics, and engineering. Systems of linear equations are among the primary objects studied in linear algebra.
Within this framework, Cramer's Rule is a practical method for solving specific systems of equations using determinants. Linear algebra techniques can handle more complex and larger systems, extending beyond two or three variable cases.
Moreover, understanding concepts such as vector spaces, matrix operations, and transformations are foundational in linear algebra. In practical terms, linear algebra helps in computing rapidly transformations and solving systems that arise in real-world applications. Mastery over these topics enables a deeper understanding of data structures and algorithms in computer science, optimization problems, and much more in the modern technological landscape.